Hi there. I have this exercise, which says:

Demonstrate that:

$\displaystyle xy''+(1-x)y'+\lambda y=0$

has a polynomial solution for some λ values.

Indicate the orthogonality relation between polynomials, the fundamental interval, and the weight function.

So I thought I should solve this using Frobenius method. I have one singular point at x=0, which is regular. I assumed a solution of the form:

$\displaystyle y(x)=\sum_0^{\infty}a_n x^{n+r}$

And then replacing in the diff. eq. I get:

$\displaystyle \sum_0^{\infty}a_n (n+r)(n+r-1) x^{n+r-1}+\sum_0^{\infty}a_n (n+r)x^{n+r-1}-\sum_0^{\infty}(n+r)a_n x^{n+r}+\lambda \sum_0^{\infty}a_n x^{n+r}=0$

$\displaystyle \sum_0^{\infty}a_n (n+r)^2 x^{n+r-1}-\sum_0^{\infty}a_n (n+r-\lambda) x^{n+r}=0$

$\displaystyle a_0r^2x^{r-1}+\sum_1^{\infty}a_n (n+r)^2 x^{n+r-1}-\sum_0^{\infty}a_n (n+r-\lambda) x^{n+r}=0$

Therefore r=0.

Then replacing r=0, and changing the index for the first summation, with m=n-1, n=m+1:

$\displaystyle \sum_0^{\infty}a_{m+1} (m+1)^2 x^{m}-\sum_0^{\infty}a_n (n-\lambda) x^{n}=0$

And now calling m=n

$\displaystyle \sum_0^{\infty}x^m \left ( a_{m+1} (m+1)^2 x^{m}-a_m (m-\lambda) \right )=0$

So I have the recurrence relation:

$\displaystyle a_{m+1}=\frac{a_m(m-\lambda)}{(m+1)^2}$

Trying some terms:

$\displaystyle a_1=-a_0\lambda$

$\displaystyle a_2=\frac{a_1(1-\lambda)}{2^2}=-\frac{a_0\lambda(1-\lambda)}{2^2}$

$\displaystyle a_3=\frac{a_2(2-\lambda)}{3^2}=-\frac{a_0\lambda(1-\lambda)(2-\lambda)}{2^23^2}$

$\displaystyle a_4=\frac{a_3(3-\lambda)}{4^2}=-\frac{a_0\lambda(1-\lambda)(2-\lambda)(3-\lambda)}{2^23^24^2}$

I'm not sure what this gives, I tried this:

$\displaystyle a_n=-\frac{a_0\lambda(n-1-\lambda)!}{(n!)^2}$

This is wrong, because the factorial in the numerator is only defined for positive values of (n-1-λ), and if n=1 I get -λ!, which wouldn't work for a_1, unless λ=0, which gives the trivial solution. But I think it works for n>1.

So I tried in a different fashion:

$\displaystyle a_1=-a_0\lambda$

$\displaystyle a_2=\frac{a_1(1-\lambda)}{2^2}=\frac{a_0\lambda(\lambda-1)}{2^2}$

$\displaystyle a_3=\frac{a_2(2-\lambda)}{3^2}=-\frac{a_0\lambda(\lambda-1)(\lambda-2)}{2^23^2}$

$\displaystyle a_4=\frac{a_3(3-\lambda)}{4^2}=\frac{a_0\lambda(\lambda-1)(\lambda-2)(\lambda-3)}{2^23^24^2}$

And now I called:

$\displaystyle a_n=a_0\frac{(-1)^n\Gamma(\lambda-n)}{(n!)^2}$

I think this is wrong too, because for example, n=1 gives $\displaystyle a_1=-a_0 \Gamma(\lambda-1)$ which doesn't fit.

Then λ-n can't be a negative integer, and the polynomials would be given by:

$\displaystyle \sum_0^{\infty}a_0\frac{(-1)^n\Gamma(\lambda-n)}{(n!)^2}x^n$

Anyway, I took the diff. eq. into it's self adjoint form:

I actually think that I didn't have to get this explicit solution. To demonstrate what the problem asks I think I should take the equation to the self adjoint form.

$\displaystyle xy''+(1-x)y'+\lambda y=0\rightarrow y''+(\frac{1}{x}-1)y'+\frac{\lambda}{x}y=0$

Multiplying by $\displaystyle r(x)=e^{\ln (x) -x}$

I get:

$\displaystyle \frac{d}{dx}\left ( e^{\ln (x) -x}\frac{dy}{dx} \right) +\lambda\frac{e^{\ln (x) -x}}{x}y=0$

This is the self adjoint form for my differential equation. Then the weight function is given by: $\displaystyle p(x)=\frac{e^{\ln (x) -x}}{x}$

I don't know how to get the fundamental interval.

Would it be ok just to work it in this last form, with out finding an explicit solution?

How can I get the fundamental interval?